I am trying to use the fact that $\lim_{n \rightarrow \infty} \sqrt{s_n} = \sqrt{s}$ where $s_n$ is the sequence of partial sums of $\sum a_n$, the same for $\sum b_n$, then I apply algebra of limits, and that is all, but I am not sure about my process.

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    $\begingroup$ Hi and welcome to Math.SE. Please update your question using mathjax. It will make it easier for people to read. $\endgroup$ – mickep Sep 14 '15 at 18:29
  • $\begingroup$ Are you assuming that $a_i, b_i$ are each nonnegative? Otherwise $\sqrt{a_nb_n}$ is a problem. $\endgroup$ – vadim123 Sep 14 '15 at 18:30
  • $\begingroup$ yes, ai,bi are nonnegative. $\endgroup$ – Iván Galeana Aguilar Sep 14 '15 at 18:33
  • $\begingroup$ Hint: $\sqrt{a_nb_n}\leq \max(a_n,b_n) \leq a_n+b_n$ $\endgroup$ – Milo Brandt Sep 14 '15 at 18:35
  • $\begingroup$ I understand you define $s_n=\sum_{j=1}^n a_j$. Then you write "the same for $b_n$"; do you here mean that you would like to set $s_n=\sum_{j=1}^n b_j$ too? what do you have in mind exactly? :-) $\endgroup$ – Math-fun Sep 14 '15 at 18:56

Hint: Use that $$\sqrt {a_n b_n} \leq \frac{a_n + b_n}{2}$$

if $a_n, b_n$ are non-negative. Use the Comparison Test.

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    $\begingroup$ Inequality of arithmetic and geometric means, well I will try. $\endgroup$ – Iván Galeana Aguilar Sep 14 '15 at 18:40
  • $\begingroup$ It is useful, but I have troubles when I try to solve the case where ai, bi they are not non-negative. $\endgroup$ – Iván Galeana Aguilar Sep 14 '15 at 19:10
  • $\begingroup$ When the sequences can be non negative, the result doesn't hold. $\endgroup$ – Kolmo Sep 14 '15 at 21:23
  • $\begingroup$ @Kolmo Something might go wrong with the square root. $\endgroup$ – Aaron Maroja Sep 14 '15 at 23:03
  • $\begingroup$ @aaron Maroja: but even if it is not an issue, the theorem can fail. $\endgroup$ – Kolmo Sep 15 '15 at 7:26

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